The average mark obtained by 22 candidates in an examination is 45. The average of the first ten is 55 while the last eleven is 40 .The marks obtained by the 11th candidate is ?

• got the answer !!
  
  Ans = 0
  
  22*45 = 990
  the avg of 1st 10 mem is 10*55= 550
  the avg of the last 11 mem is 11*40 = 440
  Sum is = 990
  11 candidate marks are Total avg - (avg of 1st 10 mem + avg of last 11 mem)
  i.e 990-990 = 0
• 11 years agoHelpfull: Yes(26) No(1)
• 0 marks
  total marks of all candidate=22*45=990
  first 10 student marks=10*55=550
  and last 11 student marks=11*40=440
  11th candidate marks=990-(550+440)=0
• 11 years agoHelpfull: Yes(4) No(0)
• total mark of 22 students = 45 *22 =990
  
  avg of 22 students =(total mark of first 10 students+mark of 11th student+total mark of last 11 students)/22
  45*22=(550+x+440)
  therefore x=0
• 11 years agoHelpfull: Yes(3) No(0)
• its 0
  because (total 22 stu marks)/22= 45 --> sum of 22 stu marks=990
  Now (total of 1st 10 stu marks)/10=55===>sum of 10 stu marks=550
  that means sum of rem 11 stu marks=440
  11th student mark= total 22 stu marks-(total 10 stu marks+remaining 11 stu)=(990-990)
• 11 years agoHelpfull: Yes(2) No(0)
• answer:40
  
  avg of 1st 10=55
  avg of last 11=40
  avg of total 22=45
  
  so,
  total avg =(mark of 1st 10 + mark of 11th + mark of last 11)/3
  =>45=(55+x+40)/3
  =>x=40
• 11 years agoHelpfull: Yes(0) No(15)